Computational complexity of solving polynomial differential equations over unbounded domains Amaury Pouly Joint work with Daniel Graça 10 May 2018 −∞ / 19
Ordinary Differential Equations (ODEs) System of ODEs: y 1 ( 0 )= y 0 , 1 y ′ 1 ( t )= f 1 ( y 1 ( t ) , . . . , y n ( t ) , t ) . . . . . . y n ( 0 )= y 0 , n y ′ n ( t )= f n ( y 1 ( t ) , . . . , y n ( t ) , t ) More compactly: y ′ ( t ) = f ( y ( t ) , t ) y ( 0 ) = y 0 1 / 19
Ordinary Differential Equations (ODEs) System of ODEs: y 1 ( 0 )= y 0 , 1 y ′ 1 ( t )= f 1 ( y 1 ( t ) , . . . , y n ( t ) , t ) . . . . . . y n ( 0 )= y 0 , n y ′ n ( t )= f n ( y 1 ( t ) , . . . , y n ( t ) , t ) More compactly: y ′ ( t ) = f ( y ( t ) , t ) y ( 0 ) = y 0 Get rid of the time: � y ( 0 ) = y 0 � y ′ ( t ) = f ( y ( t ) , z ( t )) z ( 0 ) = 0 z ′ ( t ) = 1 1 / 19
Ordinary Differential Equations (ODEs) System of ODEs: y 1 ( 0 )= y 0 , 1 y ′ 1 ( t )= f 1 ( y 1 ( t ) , . . . , y n ( t ) , t ) . . . . . . y n ( 0 )= y 0 , n y ′ n ( t )= f n ( y 1 ( t ) , . . . , y n ( t ) , t ) More compactly: y ′ ( t ) = f ( y ( t ) , t ) y ( 0 ) = y 0 Get rid of the time: � y ( 0 ) = y 0 � y ′ ( t ) = f ( y ( t ) , z ( t )) z ( 0 ) = 0 z ′ ( t ) = 1 In this talk: autonomous first order explicit system of ODEs y ′ = f ( y ) y : ( a , b ) → R n y ( 0 ) = y 0 1 / 19
A word on computability for real functions Classical computability (Turing machine): compute on words, integers, rationals, ... 2 / 19
A word on computability for real functions Classical computability (Turing machine): compute on words, integers, rationals, ... Real computability: at least two different notions BSS (Blum-Shub-Smale) machine: register machine that can store arbitrary real numbers and that can compute rational functions over reals at unit cost. Comparisons between reals are allowed. 2 / 19
A word on computability for real functions Classical computability (Turing machine): compute on words, integers, rationals, ... Real computability: at least two different notions BSS (Blum-Shub-Smale) machine: register machine that can store arbitrary real numbers and that can compute rational functions over reals at unit cost. Comparisons between reals are allowed. Computable Analysis: reals are represented as converging Cauchy sequences, computations are carried out by rational approximations using Turing machines. Comparisons between reals is not decidable in general. Computable implies continuous. 2 / 19
A word on computability for real functions Classical computability (Turing machine): compute on words, integers, rationals, ... Real computability: at least two different notions BSS (Blum-Shub-Smale) machine: register machine that can store arbitrary real numbers and that can compute rational functions over reals at unit cost. Comparisons between reals are allowed. Computable Analysis: reals are represented as converging Cauchy sequences, computations are carried out by rational approximations using Turing machines. Comparisons between reals is not decidable in general. Computable implies continuous. In this talk (unless specified) We use Computable Analysis. 2 / 19
Computability of solutions: the theory Let I = ( a , b ) and f ∈ C 0 ( R n ) . Assume y ∈ C 1 ( I , R n ) satisfies ∀ t ∈ I : y ( 0 ) = 0 , y ′ ( t ) = f ( y ( t )) . (1) Given t ∈ I and n ∈ N , can we compute q ∈ Q n s.t. � q − y ( t ) � � 2 − n ? 3 / 19
Computability of solutions: the theory Let I = ( a , b ) and f ∈ C 0 ( R n ) . Assume y ∈ C 1 ( I , R n ) satisfies ∀ t ∈ I : y ′ ( t ) = f ( y ( t )) . y ( 0 ) = 0 , (1) Is y computable? 3 / 19
Computability of solutions: the theory Let I = ( a , b ) and f ∈ C 0 ( R n ) . Assume y ∈ C 1 ( I , R n ) satisfies ∀ t ∈ I : y ′ ( t ) = f ( y ( t )) . y ( 0 ) = 0 , (1) Is y computable? Theorem (Pour-El and Richards) There exists a computable (hence continuous) f such that none of the solutions to (1) is computable. 3 / 19
Computability of solutions: the theory Let I = ( a , b ) and f ∈ C 0 ( R n ) . Assume y ∈ C 1 ( I , R n ) satisfies ∀ t ∈ I : y ′ ( t ) = f ( y ( t )) . y ( 0 ) = 0 , (1) Is y computable? Theorem (Pour-El and Richards) There exists a computable (hence continuous) f such that none of the solutions to (1) is computable. Theorem (Ruohonen) If f is computable and (1) has a unique solution, then it is computable. 3 / 19
Computability of solutions: the theory Let I = ( a , b ) and f ∈ C 0 ( R n ) . Assume y ∈ C 1 ( I , R n ) satisfies ∀ t ∈ I : y ′ ( t ) = f ( y ( t )) . y ( 0 ) = 0 , (1) Is y computable? Theorem (Pour-El and Richards) There exists a computable (hence continuous) f such that none of the solutions to (1) is computable. Theorem (Ruohonen) If f is computable and (1) has a unique solution, then it is computable. Theorem (Buescu, Campagnolo and Graça) Computing the maximum interval of life (or deciding if it is bounded) is undecidable, even if f is a polynomial. 3 / 19
Computability of solutions: the theory Let I = ( a , b ) and f ∈ C 0 ( R n ) . Assume y ∈ C 1 ( I , R n ) satisfies ∀ t ∈ I : y ′ ( t ) = f ( y ( t )) . y ( 0 ) = 0 , (1) Is y computable? Theorem (Pour-El and Richards) There exists a computable (hence continuous) f such that none of the solutions to (1) is computable. Theorem (Ruohonen) If f is computable and (1) has a unique solution, then it is computable. Theorem (Buescu, Campagnolo and Graça) Computing the maximum interval of life (or deciding if it is bounded) is undecidable, even if f is a polynomial. Theorem (Collins and Graça) The map f �→ y ( · ) for those f where y is unique, is computable. 3 / 19
Complexity of solutions: typical textbook result Assume f Lipschitz and computable, and y : I → R n satisfies ∀ t ∈ I : y ′ ( t ) = f ( y ( t )) . y ( 0 ) = 0 , 4 / 19
Complexity of solutions: typical textbook result Assume f Lipschitz and computable, and y : I → R n satisfies ∀ t ∈ I : y ′ ( t ) = f ( y ( t )) . y ( 0 ) = 0 , Theorem (Folklore, simplified) The classical Runge–Kutta method is a fourth-order method: 4 / 19
Complexity of solutions: typical textbook result Assume f Lipschitz and computable, and y : I → R n satisfies ∀ t ∈ I : y ′ ( t ) = f ( y ( t )) . y ( 0 ) = 0 , Theorem (Folklore, simplified) The classical Runge–Kutta method is a fourth-order method: given a time t ∈ I and a time step h , the algorithm returns q ∈ Q n s.t. � 1 h 4 � � � � q − y ( t ) � � O and has running time O . h 4 4 / 19
Complexity of solutions: typical textbook result Assume f Lipschitz and computable, and y : I → R n satisfies ∀ t ∈ I : y ′ ( t ) = f ( y ( t )) . y ( 0 ) = 0 , Theorem (Folklore, simplified) The classical Runge–Kutta method is a fourth-order method: given a time t ∈ I and a time step h , the algorithm returns q ∈ Q n s.t. � 1 h 4 � � � � q − y ( t ) � � O and has running time O . h 4 Usually followed by benchmarks. 4 / 19
Complexity of solutions: typical textbook result Assume f Lipschitz and computable, and y : I → R n satisfies ∀ t ∈ I : y ′ ( t ) = f ( y ( t )) . y ( 0 ) = 0 , Theorem (Folklore, simplified) The classical Runge–Kutta method is a fourth-order method: given a time t ∈ I and a time step h , the algorithm returns q ∈ Q n s.t. � 1 h 4 � � � � q − y ( t ) � � O and has running time O . h 4 Usually followed by benchmarks. Problems with this approach: � Ah 4 but A is unknown � h 4 � Accuracy of the result? O Same problem with complexity f is Lipschitz: typically only holds over compact domains 4 / 19
Complexity of solutions: typical textbook result Assume f computable and K -Lipschitz, and y : I → R n satisfies ∀ t ∈ I : y ′ ( t ) = f ( y ( t )) . y ( 0 ) = 0 , 5 / 19
Complexity of solutions: typical textbook result Assume f computable and K -Lipschitz, and y : I → R n satisfies ∀ t ∈ I : y ′ ( t ) = f ( y ( t )) . y ( 0 ) = 0 , Theorem (Folklore, simplified) Euler’s method global truncation error is: hM � e Kt − 1 � � . � � where M = sup � y ′′ ( u ) 2 K u ∈ I 5 / 19
Complexity of solutions: typical textbook result Assume f computable and K -Lipschitz, and y : I → R n satisfies ∀ t ∈ I : y ′ ( t ) = f ( y ( t )) . y ( 0 ) = 0 , Theorem (Folklore, simplified) Euler’s method global truncation error is: hM � e Kt − 1 � � . � � = O ( h ) where M = sup � y ′′ ( u ) 2 K u ∈ I In particular it has order 1 over compact time ( I ) domains. 5 / 19
Complexity of solutions: typical textbook result Assume f computable and K -Lipschitz, and y : I → R n satisfies ∀ t ∈ I : y ′ ( t ) = f ( y ( t )) . y ( 0 ) = 0 , Theorem (Folklore, simplified) Euler’s method global truncation error is: hM � e Kt − 1 � � . � � = O ( h ) where M = sup � y ′′ ( u ) 2 K u ∈ I In particular it has order 1 over compact time ( I ) domains. This bound is “useless” unless: you know K : f must be Lipschitz on “ { y ( u ) : u ∈ I } ” or globally 5 / 19
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